Prime Ring Problem （典型dfs）

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. 

Note: the number of first circle should always be 1. 

 

Input

n (0 < n < 20). 

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order. 

You are to write a program that completes above process. 

Print a blank line after each case. 

Sample Input

6
8

Sample Output

Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2


   

                        
    
                        
    
                        
    
                    
   
代码：
import java.util.Arrays;
import java.util.Scanner;
public class 素数环 {
	static int map[]=new int[25];
	static int vis[]=new int[25];
	static int n;     //i不可以随便在开头定义啊！！！！！！！！！！！！
	static int isPrime(int m){//判断素数
		if(m==0 || m==1)return 0;
		double k=Math.sqrt(m);
		int i;
		for(i=2;i<=k;i++){
			if(m%i==0)break;
		}
		if(i>k)return 1;
		else return 0;
	}
    static void dfs(int cur){
          if(cur==n){
        	  if(isPrime(map[1]+map[n])==1){//记得是环！
        	  for(int i=1;i<n;i++){
        		  System.out.print(map[i]+" ");
        	  }
        	  System.out.println(map[n]);
        	  }
        	  return;
          }
        	  for(int i=1;i<=n;i++){
        		  if(vis[i]==0 && isPrime(i+map[cur])==1){
        			  map[cur+1]=i;
        			  vis[i]=1;
        			  dfs(cur+1);
        			  vis[i]=0;
        		  }
        	  }
    }
	public static void main(String[] args) {
		// TODO Auto-generated method stub
        Scanner scan=new Scanner(System.in);
        int num=0;
        while(scan.hasNext()){
        	Arrays.fill(vis,0);
        	n=scan.nextInt();
        	System.out.println("Case "+(++num)+":");
        	map[1]=1;
        	vis[1]=1;
        	dfs(1);
        	System.out.println();
        }
	}
}

另一种判断素数：
import java.util.Arrays;
import java.util.Scanner;
public class Main {
	static int map[]=new int[25];
	static int vis[]=new int[25];
	static int prime[]=new int[1000];//标记素数数组
	static int n;
    static void dfs(int cur){
          if(cur==n){
        	  if(prime[map[1]+map[n]]==0){
        	  for(int i=1;i<=n-1;i++){
        		  System.out.print(map[i]+" ");
        	  }
        	  System.out.println(map[n]);
        	  }
        	  return;
          }
        	  for(int i=1;i<=n;i++){
        		  if(vis[i]==0 && prime[i+map[cur]]==0){
        			  map[cur+1]=i;
        			  vis[i]=1;
        			  dfs(cur+1);
        			  vis[i]=0;
        		  }
        	  }       
    }
	public static void main(String[] args) {
		// TODO Auto-generated method stub
        Scanner scan=new Scanner(System.in);
        int num=0;
        Arrays.fill(prime,0);
        prime[1]=1;//1不是素数，标记为1
        for(int i=2;i<Math.sqrt(1000);i++){
        	if(prime[i]==0){     //如果是素数，其倍数肯定不是素数
        		for(int j=i+i;j<1000;j+=i)
        			prime[j]=1;
        	}
        }
        while(scan.hasNext()){
        	Arrays.fill(vis,0);
        	n=scan.nextInt();
        	num++;
        	System.out.println("Case "+num+":");
        	map[1]=1;
        	vis[1]=1;
        	dfs(1);
        	System.out.println();
        }
	}
}